why there is no derivative in sharp turns? why there is no derivative in sharp turns in functions? I understand that it may be difficult or impossible to actually draw a tangent at that point, but is there a mathematical proof that there is no derivative in sharp turns? thanks!
 A: Suppose we have a map $f:(a,b)\rightarrow \mathbb{R}$.  Then we say that $f$ is differentiable at the point $x\in (a,b)$ if the limit 
$$\lim_{y\rightarrow x}{ \frac{f(x)-f(y)}{x-y}}$$
exists.  Now, for the situation you're looking at, the one-sided limits exist, but these limits existing does not imply that the limit exists; indeed, it only happens when these two one-sided limits exist and are equal to one another.  
For example, $f(x)=|x|$ gives a nice example, because for $x=0$ we see that the limit from the right is $1$, while the limit from the left is $-1$.  Since these two one-sided limits do not agree, the limit does not exist, and it is not differentiable.
A: Differentiable functions are locally "linear-like". Zoom in and function and tangent will be more and more similar. But zooming in a sharp turn always gives a sharp turn.
The definition of differentiability
$$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$
can be interpreted as
$$f'(c)\approx\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$
or
$$f(x)\approx f'(c)(x-c)+f(c).$$
(function is $\approx$ a straight line near $c$)
Playing with a graphing program can be very instructive.
A: The derivative at point $x_0$ exists if and only if the following limit exists:
$$\lim_{x\to x_0} \frac{f(x_0)-f(x)}{x_0-x}$$
Let's consider the example $f(x) = |x|$:
The one sided limit from below is $$\lim_{x\uparrow 0} \frac{f(0)-f(x)}{0-x} = -1$$ but the one sided limit from above is $$\lim_{x\downarrow 0} \frac{f(0)-f(x)}{0-x} = 1.$$ Note that if the (not-one-sided) limit exists, then these two limits must coincide. This means we can conclude that the above limit does not exist which means the derivative does not exists at $0$.
A: A geometric answer:  At a sharp corner, there are many possible tangent lines; any line that (locally) intersects the curve only at the corner point meets the geometric definition of a tangent.  These lines will have slopes in the closed interval between the two one-sided limits approaching the corner point. Which one do you pick as the derivative?
A: Imagine this corner, what is the tangent line? If it is flat on top(which makes the most geometric sense, it will become a straight line, if you make it a sudden decline, that works from this side, but if you look from the other side it would increase past the corner. There is no tangent line that works.
A: The geometric interpretation of the derivative is that it is the slope of the tangent line at that point. The existence of the derivative is therefore 
related to the existence of a tangent line. But just imagine, how many tangent lines can you draw at point x=0? Infinite! There isn't "a" tangent line. 
http://www.intuitive-calculus.com/limits-at-corners.html
